Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x-2y &= -1 \\ -3x+y &= 4\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-3x = -y+4$ Divide both sides by $-3$ to isolate $x$ $x = {\dfrac{1}{3}y - \dfrac{4}{3}}$ Substitute this expression for $x$ in the first equation. $8({\dfrac{1}{3}y - \dfrac{4}{3}}) - 2y = -1$ $\dfrac{8}{3}y - \dfrac{32}{3} - 2y = -1$ Simplify by combining terms, then solve for $y$ $\dfrac{2}{3}y - \dfrac{32}{3} = -1$ $\dfrac{2}{3}y = \dfrac{29}{3}$ $y = \dfrac{29}{2}$ Substitute $\dfrac{29}{2}$ for $y$ in the top equation. $8x-2( \dfrac{29}{2}) = -1$ $8x-29 = -1$ $8x = 28$ $x = \dfrac{7}{2}$ The solution is $\enspace x = \dfrac{7}{2}, \enspace y = \dfrac{29}{2}$.